Cow Contest

?POJ - 3660?

N?(1 ≤?N?≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

The contest is conducted in several head-to-head rounds, each between two cows. If cow?A?has a greater skill level than cow?B?(1 ≤?A?≤?N; 1 ≤?B?≤?N;?A?≠?B), then cow?A?will always beat cow?B.

Farmer John is trying to rank the cows by skill level. Given a list the results of?M(1 ≤?M?≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.

Input

* Line 1: Two space-separated integers:?N?and?M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer,?A, is the winner) of a single round of competition:?A?and?B

Output

* Line 1: A single integer representing the number of cows whose ranks can be determined
　

Sample Input

5 5 4 3 4 2 3 2 1 2 2 5

Sample Output

2

題目大意：第一行給出兩個數n,m，n代表奶牛的數量，m代表m場比賽，下面m行每行有兩個整數a,b，代表本場比賽a能夠擊敗b，（注：若a能擊敗b，b能擊敗c，則代表a能擊敗c）問最后能確定多少頭奶牛的排名。

解決方法：通過使用floyd算法，將每兩頭奶牛的勝負關系確定，如果知道一頭奶牛與其他所有奶牛的勝負關系，則可推出這頭奶牛的排名。

AC代碼：

#include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstdlib> #include <cstring> #include <map> #include <stack> #include <queue> #include <vector> #include <bitset> #include <set> #include <utility> using namespace std; typedef long long ll; #define inf 0x3f3f3f3f #define rep(i,l,r) for(int i=l;i<=r;i++) #define lep(i,l,r) for(int i=l;i>=r;i--) #define ms(arr) memset(arr,0,sizeof(arr)) //priority_queue<int,vector<int> ,greater<int> >q; const int maxn = (int)1e3 + 5; const ll mod = 1e9+7; bool mapp[maxn][maxn]; int num[maxn]; int n,m; void floyd() {rep(k,1,n) {rep(i,1,n) {rep(j,1,n) {if(mapp[i][k]==true&&mapp[k][j]==true)mapp[i][j]=true;}}} } int main() {//freopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout);ios::sync_with_stdio(0),cin.tie(0);cin>>n>>m;memset(mapp,false,sizeof(mapp));while(m--){int a,b;cin>>a>>b;mapp[a][b]=true;}floyd();int ans=0;rep(i,1,n) {rep(j,1,n) {if(mapp[i][j]==true||mapp[j][i]==true)num[i]++;}if(num[i]==n-1)ans++;}cout<<ans<<endl;return 0; }

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